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Difference between revisions of "Distance formula"
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Shortest distance from a point to a line:
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the distance
 +
between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is
 +
 +
    <math>|ax_1+by_1+c|/\sqrt(a^2+b^2)</math>
 +
 +
            Proof:
 +
The equation <math>ax + by + c = 0</math> can be written:
 +
 +
    <math>y = -(a/b)x - (c/a)</math>
 +
 +
So the perpendicular line through (x1,y1) is:
 +
 +
    <math>x-x_1</math>  <math>y-y_1</math>
 +
    ----  = ---- =  <math>t/\sqrt(a^2+b^2)</math>    where t is a parameter.
 +
      a        b
 +
 +
t will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to (x,y).
 +
 +
So
 +
 +
    <math>x = x_1 + a \dot t/\sqrt(a^2+b^2)</math>
 +
 +
and
 +
 +
    <math>y = y_1 + b \dot t/\sqrt(a^2+b^2)</math>
 +
 +
This meets the given line ax+by+c = 0 where:
 +
 +
    a(x1+a.t/sqrt(a^2+b^2)) + b(y1+b.t/sqrt(a^2+b^2)) + c = 0
 +
 +
              ax1 + by1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0   
 +
 +
                          ax1 + by1 + c + t.sqrt(a^2+b^2) = 0
 +
 +
so
 +
 +
    t.sqrt(a^2+b^2) = -(ax1+by1+c)
 +
 +
                  t = -(ax1+by1+c)/sqrt(a^2+b^2)
 +
 +
Therefore the perpendicular distance from (x1,y1) to the line
 +
ax+by+c = 0 is:
 +
 +
            ax1 + by1 + c
 +
    |t| =  -------------
 +
            sqrt(a^2+b^2)

Revision as of 12:35, 3 April 2011

The distance formula is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. In the two-dimensional case, it says that the distance between two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ is given by $d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$. In the $n$-dimensional case, the distance between $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ is $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}$


This article is a stub. Help us out by expanding it.

Shortest distance from a point to a line: the distance between the line $ax+by+c = 0$ and point $(x_1,y_1)$ is 

    $|ax_1+by_1+c|/\sqrt(a^2+b^2)$

            Proof:

The equation $ax + by + c = 0$ can be written: 

    $y = -(a/b)x - (c/a)$

So the perpendicular line through (x1,y1) is: 

   $x-x_1$   $y-y_1$
    ----   = ---- =  $t/\sqrt(a^2+b^2)$     where t is a parameter.
     a         b

t will be the distance from the point $(x_1,y_1)$ along the perpendicular line to (x,y).

So 

    $x = x_1 + a \dot t/\sqrt(a^2+b^2)$

and 

    $y = y_1 + b \dot t/\sqrt(a^2+b^2)$

This meets the given line ax+by+c = 0 where: 

    a(x1+a.t/sqrt(a^2+b^2)) + b(y1+b.t/sqrt(a^2+b^2)) + c = 0

             ax1 + by1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0    

                          ax1 + by1 + c + t.sqrt(a^2+b^2) = 0

so 

    t.sqrt(a^2+b^2) = -(ax1+by1+c)

                  t = -(ax1+by1+c)/sqrt(a^2+b^2)

Therefore the perpendicular distance from (x1,y1) to the line ax+by+c = 0 is: 

           ax1 + by1 + c
    |t| =  ------------- 
           sqrt(a^2+b^2)
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